Question

The energy stored in the capacitor (in $\mu J$ ) between terminals $a$ and $b$ of the network (as shown in figure) is $\frac{n}{2}\mu J$ . Fin the value of $n$ . If capacitance of each capacitor is $1\mu F$ .

Answer 1

Answer: 25

Redrawing network, we get a balanced Wheatstone's network.

$\Rightarrow C_{eq}=C$
Charge on capacitor between terminals $a$ and $b$
$\frac{Q}{2}=\frac{CV}{2}$
Energy stored in that capacitor $=\frac{1}{2}\frac{{\left(\frac{Q}{2}\right)}^2}{C}$
$\begin{array}{l}=\frac{Q^2}{8C}\\ =\frac{C{V}^2}{8}\\ =\frac{1\times 10^{-6}\times 10^2}{8}J\\ =12.5\mu J\end{array}$